In a certain region of space, there exists a uniform and constant electric field of strength E along x-axis and uniform constant magnetic field of induction B along z-axis. A charge particle having charge q and mass m is projected with speed v parallel to x-axis from a point (a, b, 0). When the particle reaches a point( 2a, `b//2`, 0) its speed becomes 2v. Find the value of electric field strength in term of m, v and co-ordinates.

Options

A. `(3)/(2)(mv^(2))/(qa)`
B. `(mv^(2))/(qb)`
C. `(2mv^(2))/(qa)`
D. `(mv^(2))/(2qa)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Forces Acting on the Particle
The charged particle experiences two forces:
1. The electric force due to the electric field \( E \) along the x-axis, given by \( F_E = qE \).
2. The magnetic force due to the magnetic field \( B \) along the z-axis, which does not perform work since it is always perpendicular to the velocity of the particle.

### Step 2: Work Done by the Electric Field
Since the magnetic force does not do any work, the work done on the particle is entirely due to the electric field. The work done \( W \) by the electric field when the particle moves from \( (a, b, 0) \) to \( (2a, \frac{b}{2}, 0) \) can be calculated as:
\[ W = F_E \cdot d \]
where \( d \) is the displacement in the direction of the electric field. The displacement along the x-axis is:
\[ d = 2a - a = a \]
Thus, the work done by the electric field is:
\[ W = qE \cdot a \]

### Step 3: Change in Kinetic Energy
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE \]
The initial kinetic energy \( KE_i \) when the particle is projected with speed \( v \) is:
\[ KE_i = \frac{1}{2} mv^2 \]
The final kinetic energy \( KE_f \) when the particle reaches the point \( (2a, \frac{b}{2}, 0) \) with speed \( 2v \) is:
\[ KE_f = \frac{1}{2} m(2v)^2 = \frac{1}{2} m(4v^2) = 2mv^2 \]
The change in kinetic energy is:
\[ \Delta KE = KE_f - KE_i = 2mv^2 - \frac{1}{2} mv^2 = \frac{4mv^2}{2} - \frac{1}{2} mv^2 = \frac{3}{2} mv^2 \]

### Step 4: Equate Work Done to Change in Kinetic Energy
Now, we can equate the work done by the electric field to the change in kinetic energy:
\[ qE \cdot a = \frac{3}{2} mv^2 \]

### Step 5: Solve for Electric Field Strength \( E \)
Rearranging the equation to solve for \( E \):
\[ E = \frac{\frac{3}{2} mv^2}{qa} \]

### Final Result
Thus, the electric field strength \( E \) in terms of \( m \), \( v \), \( q \), and \( a \) is:
\[ E = \frac{3mv^2}{2qa} \]